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Whole roots and powers

  1. Nov 28, 2011 #1

    georg gill

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    [tex](a^{\frac{1}{2n}}a^{\frac{1}{2n}})^n=(a^{\frac{1}{2n}})^n(a^{\frac{1}{2n}})^n=(a^{\frac{1}{2}}) (a^{\frac{1}{2}})=a=(a^{\frac{1}{n}})^n[/tex]

    all above is just done by using that the order of the factors that you multiply does not matter

    we have proven that

    [tex](a^{\frac{1}{p}}a^{\frac{1}{p}})^n=(a^{\frac{2}{p}})^n[/tex]

    for any even number. Try saying that p is a odd number then n is a decimal number and taking the rooth with a number that is not whole does not make sense in a logixal approach of what one can really comprehend. For any even number of p what I did above makes sense. Could someone see an extention of this system. I want to make left side here:



    [tex]\sqrt[m]{\underbrace{a^{\frac{1}{n}} \, \cdot \, a^{\frac{1}{n}} \, \cdot \, . . . \, \cdot a^{\frac{1}{n}} \, }_{\text{m times}} \, }=a^{\frac{m}{n}} [/tex]

    to become right side by only using roths and powers of integers.
     
    Last edited: Nov 28, 2011
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  3. Nov 28, 2011 #2

    HallsofIvy

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    Certainly [itex]a^{1/2n}a^{1/2n}= a^{1/2n+ 1/2n}= a^{2/2n}= a^{1/n}[/itex].

    And [itex]\left(a^{1/n}\right)^m= a^{(1/n)m}= a^{m/n}[/itex]. Those are well known properties of the exponentials.
     
    Last edited: Nov 28, 2011
  4. Nov 28, 2011 #3

    D H

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    If you know what Dedekind cuts are, the PlanetMath page http://planetmath.org/encyclopedia/ProofOfPropertiesOfTheExponential.html [Broken] does a nice job of extending exponentiation from the integers to the rationals, and then to the reals.
     
    Last edited by a moderator: May 5, 2017
  5. Nov 29, 2011 #4

    georg gill

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    As far as I can see or from what I get from the proof at least

    [tex](xy)^m)^{\frac{1}{n}}=(y)^m)^{\frac{1}{n}}(x)^m)^{\frac{1}{n}}[/tex]

    I want to prove that

    [tex](xy)^m)^{\frac{1}{n}}=y)^{\frac{m}{n}}x^{\frac{m}{n}}[/tex]

    What I have managed to show is that

    [tex](xy^{\frac{1}{np}})^{np}[/tex]

    [tex](a^{\frac{1}{np}})^{np}=a[/tex]

    [tex]((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p[/tex]

    since n and p are integers to put p outside as own power goues without proving (could of course prove this
    and we get

    [tex]((a^{\frac{1}{n}})^{\frac{1}{p}})^{np}=(((a^{\frac{1}{n}})^{\frac{1}{p}})^n)^p=a[/tex]

    so

    [tex]xy^{\frac{1}{np}}=(a^{\frac{1}{n}})^{\frac{1}{p}}[/tex]
     
    Last edited by a moderator: May 5, 2017
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